Question: Find the shortest distance between the point $(6,12)$ and the parabola given by the equation $x = \frac{y^2}{2}.$
Solution: Let $P = \left( \frac{a^2}{2}, a \right)$ be a point on the parabola.  First, we find the equation of the tangent to the parabola at $P.$

[asy]
unitsize(0.5 cm);

real y;

pair P = (8,4);
path parab = ((-5)^2/2,-5);

for (y = -5; y <= 5; y = y + 0.01) {
  parab = parab--(y^2/2,y);
}

draw(parab,red);
draw((P + (-4,-4/4))--(P + (4,4/4)),dashed);

draw((-2,0)--(15,0));
draw((0,-5)--(0,5));

dot("$P$", P, S);
[/asy]

Since the tangent passes through $\left( \frac{a^2}{2}, a \right),$ the equation of the tangent is of the form
\[y - a = m \left( x - \frac{a^2}{2} \right) = mx - \frac{a^2 m}{2}.\]Substituting $x = \frac{y^2}{2},$ we get
\[y - a = \frac{my^2}{2} - \frac{a^2 m}{2}.\]This simplifies to $my^2 - 2y + 2a - a^2 m = 0.$  Since this is the equation of a tangent, the quadratic should have a double root of $y = a,$ which means its discriminant is 0, which gives us
\[4 - 4m(2a - a^2 m) = 0.\]Then $4a^2 m^2 - 8am + 4 = 4(am - 1)^2 = 0,$ so $m = \frac{1}{a}.$

Now, consider the point $P$ that is closest to $(6,12).$

[asy]
unitsize(0.5 cm);

real y;

pair P = (8,4);
path parab = ((-2)^2/2,-2);

for (y = -2; y <= 5; y = y + 0.01) {
  parab = parab--(y^2/2,y);
}

draw(parab,red);

draw((-2,0)--(15,0));
draw((0,-2)--(0,15));
draw(P--(6,12));
draw((P + (-4,-4/4))--(P + (4,4/4)),dashed);

dot("$(6,12)$", (6,12), N);
dot("$P$", P, S);
[/asy]

Geometrically, the line connecting $P$ and $(6,12)$ is perpendicular to the tangent.  In terms of slopes, this gives us
\[\frac{a - 12}{\frac{a^2}{2} - 6} \cdot \frac{1}{a} = -1.\]This simplifies to $a^3 - 10a - 24 = 0,$ which factors as $(a - 4)(a^2 + 4a + 6) = 0.$  The quadratic factor has no real roots, so $a = 4.$  Therefore, $P = (8,4),$ and the shortest distance is $\sqrt{(8 - 6)^2 + (4 - 12)^2} = \boxed{2 \sqrt{17}}.$